What is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?
Only 23 people need to be in the room.
Our first observation in solving this problem is the following:
(the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0
What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations).
With some simple re-arranging of the formula, we get:
the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday)
So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer.
The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far.
These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918
More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365)
We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%.
Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.
Allan, Bertrand, and Cecil were caught stealing so the king sent them to the dungeon.
But the king decided to give them a chance.
He mad them stand in a line and put hats on their heads.
He told them that if they answer a riddle, they could go free.
Here is the riddle: "Each of you has a hat on your head. You do not know the color of the hat on your own head. If one of you can guess the color of the hat on your head, I will let you free. But before you answer you must keep standing in this line. You cannot turn around. Here are my only hints: there are only black and white hats. At least one hat is black. At least one hat is white."
Allan couldn't see any hats.
Bertrand could see Allan's hat but not his own.
Cecil could see Bertrand's hat and Allan's hat, but not his own.
After a minute nobody had solved the riddle. But then a short while later, one of them solved the riddle. Who was is and how did he know?
Bertrand knew the answer because Cecil didn't say anything after one minute. If Bertrand and Allan's hats were both the same color, then Cecil would know what color his hat was. But Cecil didn't know. So Bertrand knew that Allan's hat was a different color than his. Since Allan's hat was black, Betrand knew his hat was white.
One morning an airline president is leaving on a business trip and finds he left some paperwork at his office. He runs into his office to get it and the night watchman stops him and says, "Sir, don't get on the plane. I had a dream last night that the plane would crash and everyone would die!"
The man takes his word and cancels his trip. Sure enough, the plane crashes and everyone dies. The next morning the man gives the watchman a $1,000 reward for saving his life and then fires him.
Why did he fire the watchman that saved his life?
You are standing in a house in the middle of the countryside. There is a small hole in one of the interior walls of the house, through which 100 identical wires are protruding.
From this hole, the wires run underground all the way to a small shed exactly 1 mile away from the house, and are protruding from one of the shed's walls so that they are accessible from inside the shed.
The ends of the wires coming out of the house wall each have a small tag on them, labeled with each number from 1 to 100 (so one of the wires is labeled "1", one is labeled "2", and so on, all the way through "100"). Your task is to label the ends of the wires protruding from the shed wall with the same number as the other end of the wire from the house (so, for example, the wire with its end labeled "47" in the house should have its other end in the shed labeled "47" as well).
To help you label the ends of the wires in the shed, there are an unlimited supply of batteries in the house, and a single lightbulb in the shed. The way it works is that in the house, you can take any two wires and attach them to a single battery. If you then go to the shed and touch those two wires to the lightbulb, it will light up. The lightbulb will only light up if you touch it to two wires that are attached to the same battery. You can use as many of the batteries as you want, but you cannot attach any given wire to more than one battery at a time. Also, you cannot attach more than two wires to a given battery at one time. (Basically, each battery you use will have exactly two wires attached to it). Note that you don't have to attach all of the wires to batteries if you don't want to.
Your goal, starting in the house, is to travel as little distance as possible in order to label all of the wires in the shed.
You tell a few friends about the task at hand.
"That will require you to travel 15 miles!" of of them exclaims.
"Pish posh," yells another. "You'll only have to travel 5 miles!"
"That's nonsense," a third replies. "You can do it in 3 miles!"
Which of your friends is correct? And what strategy would you use to travel that number of miles to label all of the wires in the shed?
Believe it or not, you can do it travelling only 3 miles!
The answer is rather elegant. Starting from the house, don't attach wires 1 and 2 to any batteries, but for the remaining wires, attach them in consecutive pairs to batteries (so attach wires 3 and 4 to the same battery, attach wires 5 and 6 to the same battery, and so on all the way through wires 99 and 100).
Now travel 1 mile to the shed, and using the lightbulb, find all pairs of wires that light it up. Put a rubberband around each pair or wires that light up the lightbulb. The two wires that don't light up any lightbulbs are wires 1 and 2 (though you don't know yet which one of them is wire 1 and which is wire 2). Put a rubberband around this pair of wires as well, but mark it so you remember that they are wires 1 and 2.
Now go 1 mile back to the house, and attach odd-numbered wires to batteries in the following pairs: (1 and 3), (5 and 7), (9 and 11), and so on, all the way through (97 and 99).
Similarly, attach even-numbered wires to batteries in the following pairs: (4 and 6), (8 and 10), (12 and 14), and so on, all the way through (96 and 98).
Note that in this round, we didn't attach wire 2 or wire 100 to any batteries.
Finally, travel 1 mile back to the shed. You're now in a position to label all of the wires here.
First, remember we know the pair of wires that are, collectively, wires 1 and 2. So test wires 1 and 2 with all the other wires to see what pair lights up the lightbulb. The wire from wires 1 and 2 that doesn't light up the bulb is wire 2 (which, remember, we didn't connect to a battery), and the other is wire 1, so we can label these as such. Furthermore, the wire that, with wire 1, lights up a lightbulb, is wire 3 (remember how we connected the wires this round).
Now, the other wire in the rubber band with wire 3 is wire 4 (we know this from the first round), and the wire that, with wire 4, lights up the lightbulb, is wire 6 (again, because of how we connected the wires to batteries this round). We can continue labeling batteries this way (next we'll label wire 7, which is rubber-banded to wire 6, and then we'll label wire 9, which lights up the lightbulb with wire 7, and so on). At the end, we'll label wire 97, and then wire 99 (which lights up the lightbulb with wire 97), and finally wire 100 (which isn't connected to a battery this round, but is rubber-banded to wire 99).
And we're done, having travelled only 3 miles!
A new student met the Zen Master after traveling hundreds of miles by yak cart. He was understandably pleased with himself for being selected to learn at the great master's feet .
The first time they formally met, the Zen Master asked, "May I ask you a simple question?" "It would be an honor!" replied the student.
"Which is greater, that which has no beginning or that which has no end?" queried the Zen Master. "Come back when you have the answer and can explain why."
After the student made many frustrated trips back with answers which the master quickly cast off with a disapproving negative nod, the Zen Master finally said, "Perhaps I should ask you another question?"
"Oh, please do!" pleaded the exasperated student.
The Zen Master then asked, "Since you do not know that, answer this much simpler riddle. When can a pebble hold back the sea?" Again the student was rebuffed time and again. Several more questions followed with the same result. Each time, the student could not find the correct answer. Finally, completely exasperated, the student began to weep, "Master, I am a complete idiot. I can not solve even the simplest riddle from you!"
Suddenly, the student stopped, sat down, and said, "I am ready for my second lesson."
What was the Zen Master's first lesson?
The student's first lesson was that in order to learn from the Zen Master, the student should be asking the questions and not the Zen Master.
A man needs to send important documents to his friend across the country. He buys a suitcase to put the documents in, but he has a problem: the mail system in his country is very corrupt, and he knows that if he doesn't lock the suitcase, it will be opened by the post office and his documents will be stolen before they reach his friend.
There are lock stores across the country that sell locks with keys. The only problem is that if he locks the suitcase, he has no way to send the key to his friend so that the friend will be able to open the lock: if he doesn't send the key, then the friend can't open the lock, and if he puts the key in the suitcase, then the friend won't be able to get to the key.
The suitcase is designed so that any number of locks can be put on it, but the man figures that putting more than one lock on the suitcase will only compound the problem.
After a few days, however, he figures out how to safely send the documents. He calls his friend who he's sending the documents to and explains the plan.
What is the man's plan?
The plan is this:
1. The man will put a lock on the suitcase, keep the key, and send the suitcase to his friend.
2. The friend will then put his own lock on the suitcase as well, keep the key to that lock, and send the suitcase back to the man.
3. The man will use his key to remove his lock from the suitcase, and send it back to the friend.
4. The friend will remove his own lock from the suitcase and get to the documents.
Search: Man-in-the-middle attack
Hussey has been caught stealing goats, and is brought into court for justice. The judge is his ex-wife Amy Hussey, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Hussey from close range.
To make things a little better for Hussey, Amy Hussey tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot.
If Hussey is still alive, she will then either take another shot, or spin the chamber again before shooting. Hussey is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower.
He steels himself as Amy Hussey loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Amy Hussey asks, 'Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?'
What should Hussey choose?
Hussey should have Amy Hussey pull the trigger again without spinning.
We know that the first chamber Amy Hussey fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Hussey has Amy Hussey pull the trigger again, the probability that a bullet will be fired is 1/4.
If Amy Hussey spins the chamber again, the probability that she shoots Hussey would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.
There are 1 million closed school lockers in a row, labeled 1 through 1,000,000.
You first go through and flip every locker open.
Then you go through and flip every other locker (locker 2, 4, 6, etc...). When you're done, all the even-numbered lockers are closed.
You then go through and flip every third locker (3, 6, 9, etc...). "Flipping" mean you open it if it's closed, and close it if it's open. For example, as you go through this time, you close locker 3 (because it was still open after the previous run through), but you open locker 6, since you had closed it in the previous run through.
Then you go through and flip every fourth locker (4, 8, 12, etc...), then every fifth locker (5, 10, 15, etc...), then every sixth locker (6, 12, 18, etc...) and so on. At the end, you're going through and flipping every 999,998th locker (which is just locker 999,998), then every 999,999th locker (which is just locker 999,999), and finally, every 1,000,000th locker (which is just locker 1,000,000).
At the end of this, is locker 1,000,000 open or closed?
Locker 1,000,000 will be open.
If you think about it, the number of times that each locker is flipped is equal to the number of factors it has. For example, locker 12 has factors 1, 2, 3, 4, 6, and 12, and will thus be flipped 6 times (it will end be flipped when you flip every one, every 2nd, every 3rd, every 4th, every 6th, and every 12th locker). It will end up closed, since flipping an even number of times will return it to its starting position. You can see that if a locker number has an even number of factors, it will end up closed. If it has an odd number of factors, it will end up open.
As it turns out, the only types of numbers that have an odd number of factors are squares. This is because factors come in pairs, and for squares, one of those pairs is the square root, which is duplicated and thus doesn't count twice as a factor. For example, 12's factors are 1 x 12, 2 x 6, and 3 x 4 (6 total factors). On the other hand, 16's factors are 1 x 16, 2 x 8, and 4 x 4 (5 total factors).
So lockers 1, 4, 9, 16, 25, etc... will all be open. Since 1,000,000 is a square number (1000 x 1000), it will be open as well.